Question

Question 4: Show that the following four conditions are equivalent: (i) A ⊂ B (ii) A B = Φ (iii) A ∪ B = B(iv) A ∩ B = A

Answer 1

let A={1,2,3,4}
B={1,2,3,4,5,6,7,8,9,0}
1) All the elements in A is also present in B
2)A-B=null set as all common elements from A will be removed
3)it is B because we add A and B
     A is a subset of B then the sum is B
4)it is A because we are writing down all the common terms when A is a subset of B then all the terms in A are common terms

Answer 2

Answer:

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Step-by-step explanation:

4. Show that the following four conditions are equivalent:

(i) A ⊂ B (ii) A – B = Φ

(iii) A ∪ B = B (iv) A ∩ B = A

Solution:

According to the question,

To prove, (i) ⬌ (ii)

Here, (i) = A ⊂ B and (ii) = A – B ≠ ϕ

Let us assume that A ⊂ B

To prove, A – B ≠ ϕ

Let A – B ≠ ϕ

Hence, there exists X ∈ A, X ≠ B, but since A⊂ B, it is not possible

∴ A – B = ϕ

And A⊂ B ⇒ A – B ≠ ϕ

Let us assume that A – B ≠ ϕ

To prove: A ⊂ B

Let X∈ A

So, X ∈ B (if X ∉ B, then A – B ≠ ϕ)

Hence, A – B = ϕ => A ⊂ B

∴(i) ⬌ (ii)

Let us assume that A ⊂ B

To prove, A ∪ B = B

⇒ B ⊂ A ∪ B

Let us assume that, x ∈ A∪ B

⇒ X ∈ A or X ∈ B

Taking Case I: X ∈ B

A ∪ B = B

Taking Case II: X ∈ A

⇒ X ∈ B (A ⊂ B)

⇒ A ∪ B ⊂ B

Let A ∪ B = B

Let us assume that X ∈ A

⇒ X ∈ A ∪ B (A ⊂ A ∪ B)

⇒ X ∈ B (A ∪ B = B)

∴A⊂ B

Hence, (i) ⬌ (iii)

To prove (i) ⬌ (iv)

Let us assume that A ⊂ B

A ∩ B ⊂ A

Let X ∈ A

To prove, X ∈ A∩ B

Since, A ⊂ B and X ∈ B

Hence, X ∈ A ∩ B

⇒ A ⊂ A ∩ B

⇒ A = A ∩ B

Let us assume that A ∩ B = A

Let X ∈ A

⇒ X ∈ A ∩ B

⇒ X ∈ B and X ∈ A

⇒ A ⊂ B

∴ (i) ⬌ (iv)

∴ (i) ⬌ (ii) ⬌ (iii) ⬌ (iv)

Hence, proved.

Bye!